calcul de l′ angle x △ABD ((sin y)/(AD))=((sin44 )/(BD)) (1) △ADC ((sin x)/(AD))=((sin48 )/(DC)) (2) (1) BD sin y=ADsin 44 (2) DCsinx =ADsin 48 (((1))/((2)))⇔ ((BDsin y)/(DCsin x))=((sin 44)/(sin 48)) (3) △BDC ((sin 14)/(BD))=((sin 28)/(DC)) ((BD)/(DC))=((sin 14)/(sin 28)) (3)⇔((sin 14×)/(sin 28×))((sin y)/(sin x))=((sin 44)/(sin 48)) Relation entre x et y △ABC (∡A +∡B+ ∡C)=180 48+44+y+28+14+x=180 x+y=46 ⇒ y=46−x ((sin x)/(sin y))=((sin 48×sin 28)/(sin 44×sin 14)) ((sinx)/(sin(46−x)))=((2sin 48×cos 14)/(sin 44)) =((sin x)/(sin 46cos x−cos 46sin x)) =((sin x)/(cos x(sin 46−tan x×cos 46))) =((tan x)/(sin 46−cos 46×tan x)) sin 44×tan x=2sin 48×cos 14(sin 46−cos 46tan x) (sin 44+2sin 48cos 14×cos 46)tan x= 2sin 48cos 14×sin 46 tan x=((2sin 48×cos 14×sin 46)/(sin 44+2sin 48×cos 14×cos 46)) sin 44= sin 48= sin 46 cos 14= cos 46= tan x=0,61150462 soit: X=31,446^°


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Question Number 185169 by a.lgnaoui last updated on 18/Jan/23

$c a l c u l d e l^{'} a n g l e x$ $△ A B D \frac{\sin y}{AD} = \frac{\sin 44}{BD} (1)$ $△ A D C \frac{\sin x}{AD} = \frac{\sin 48}{DC} (2)$ $(1) BD \sin y = ADsin 44$ $(2) DCsinx = ADsin 48$ $\frac{(1)}{(2)} \Leftrightarrow \frac{BDsin y}{DCsin x} = \frac{\sin 44}{\sin 48} (3)$ $△ B D C \frac{\sin 14}{BD} = \frac{\sin 28}{DC}$ $\frac{BD}{DC} = \frac{\sin 14}{\sin 28}$ $(3) \Leftrightarrow \frac{\sin 14 \times}{\sin 28 \times} \frac{\sin y}{\sin x} = \frac{\sin 44}{\sin 48}$ $R e l a t i o n e n t r e x e t y$ $△ A B C (∡ A + ∡ B + ∡ C) = 180$ $48 + 44 + y + 28 + 14 + x = 180$ $x + y = 46 \Rightarrow y = 46 - x$ $\frac{\sin x}{\sin y} = \frac{\sin 48 \times \sin 28}{\sin 44 \times \sin 14}$ $\frac{sinx}{\sin (46 - x)} = \frac{2 \sin 48 \times \cos 14}{\sin 44}$ $= \frac{\sin x}{\sin 46 \cos x - \cos 46 \sin x}$ $= \frac{\sin x}{\cos x (\sin 46 - \tan x \times \cos 46)}$ $= \frac{\tan x}{\sin 46 - \cos 46 \times \tan x}$ $\sin 44 \times \tan x = 2 \sin 48 \times \cos 14 (\sin 46 - \cos 46 \tan x)$ $(\sin 44 + 2 \sin 48 \cos 14 \times \cos 46) \tan x =$ $2 \sin 48 \cos 14 \times \sin 46$ $\tan x = \frac{2 \sin 48 \times \cos 14 \times \sin 46}{\sin 44 + 2 \sin 48 \times \cos 14 \times \cos 46}$ $\sin 44 = \sin 48 = \sin 46$ $\cos 14 = \cos 46 =$ $\tan x = 0, 61150462$ $s o i t : X = 31, 446^{°}$
Commented by a.lgnaoui last updated on 18/Jan/23

	Answered by a.lgnaoui last updated on 18/Jan/23

	Commented by a.lgnaoui last updated on 18/Jan/23

	$\sin 44 = 0, 694658 \sin 48 = 0, 7431448$ $\cos 14 = 0, 9702957 \cos 46 = 0, 694658$ $\sin 46 = 0, 7193398$
	Commented by a.lgnaoui last updated on 18/Jan/23

	$R e c a p i t u l a t i o n$ $- - - - - - - - - - -$ $∡ A C D = x = 31, 446$ $∡ C D A = 100, 55$ $∡ A D B = 121, 45$ $∡ A B D = y = 14, 55$ $∡ B D C = 138$
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